If your not too busy...
1-tan^2X =1-2tan^2X
1+tan^2X
First, use long division on the left hand side.
You should find
$\displaystyle \frac{1 - \tan^2{x}}{1 + \tan^2{x}} = -1 + \frac{2}{\tan^2{x} + 1}$
Remember that $\displaystyle \tan^2{x} + 1 = \sec^2{x}$
So $\displaystyle \frac{1 - \tan^2{x}}{1 + \tan^2{x}} = -1 + \frac{2}{\sec^2{x}}$
$\displaystyle = -1 + 2\cos^2{x}$
$\displaystyle = -1 + 2(1 - \sin^2{x})$, since $\displaystyle \cos^2{x} + \sin^2{x} = 1$...
$\displaystyle = -1 + 2 - 2\sin^2{x}$
$\displaystyle = 1 - 2\sin^2{x}$.
Hello Dominc O'L
$\displaystyle \frac{1-\tan^2x}{1+\tan^2x}$
$\displaystyle =\frac{1-\tan^2x}{1+\tan^2x}\cdot \frac{\cos^2x}{\cos^2x}$
$\displaystyle =\frac{\cos^2x - \sin^2x}{\cos^2x + \sin^2x}$
$\displaystyle = \frac{(1-\sin^2x)-\sin^2x}{1}$
$\displaystyle = 1 - 2\sin^2x$
Grandad